1. C – 1 Linear Programming

2. C – 2 Linear Programming  A mathematical technique to help plan and make decisions relative to the trade-offs necessary to allocate resources  Will find the minimum or maximum value of the objective  Guarantees the optimal solution to the model formulated

3. C – 3 Requirements of an LP Problem 1.LP problems seek to maximize or minimize some quantity (usually profit or cost) expressed as an objective function 2.The presence of restrictions, or constraints, limits the degree to which we can pursue our objective

4. C – 4 Requirements of an LP Problem 3.There must be alternative courses of action to choose from 4.The objective and constraints in linear programming problems must be expressed in terms of linear equations or inequalities

5. C – 5 Formulating LP Problems The product-mix problem at Shader Electronics  Two products 1.Shader Walkman, a portable CD/DVD player 2.Shader Watch-TV, a wristwatch-size Internet-connected color TV  Determine the mix of products that will produce the maximum profit

6. C – 6 Formulating LP Problems WalkmanWatch-TVsAvailable Hours Department(X 1 )(X 2 )This Week Hours Required to Produce 1 Unit Electronic43240 Assembly21100 Profit per unit$7$5 Decision Variables: X 1 = number of Walkmans to be produced X 2 = number of Watch-TVs to be produced

7. C – 7 Formulating LP Problems Objective Function: Maximize Profit = $7X 1 + $5X 2 There are three types of constraints   Upper limits where the amount used is ≤ the amount of a resource   Lower limits where the amount used is ≥ the amount of the resource   Equalities where the amount used is = the amount of the resource

8. C – 8 Formulating LP Problems Second Constraint: 2X 1 + 1X 2 ≤ 100 (hours of assembly time) Assembly time available Assembly time used is ≤ First Constraint: 4X 1 + 3X 2 ≤ 240 (hours of electronic time) Electronic time available Electronic time used is ≤

9. C – 9 Graphical Solution  Can be used when there are two decision variables 1.Plot the constraint equations at their limits by converting each equation to an equality 2.Identify the feasible solution space 3.Create an iso-profit line based on the objective function 4.Move this line outwards until the optimal point is identified

10. C – 10 Graphical Solution 100 100– – 80 80 – – 60 60 – – 40 40 – – 20 20 – – ||||||||||| 020406080100 Number of Watch-TVs Number of Walkmans X1X1X1X1 X2X2X2X2 Assembly (constraint B) Electronics (constraint A) Feasible region

11. C – 11 Graphical Solution 100 100– – 80 80 – – 60 60 – – 40 40 – – 20 20 – – ||||||||||| 020406080100 Number of Watch-TVs Number of Walkmans X1X1X1X1 X2X2X2X2 (0, 42) (30, 0) $210 = $7X 1 + $5X 2

12. C – 12 Graphical Solution 100 100– – 80 80 – – 60 60 – – 40 40 – – 20 20 – – ||||||||||| 020406080100 Number of Watch-TVs Number of Walkmans X1X1X1X1 X2X2X2X2 $210 = $7X 1 + $5X 2 $350 = $7X 1 + $5X 2 $420 = $7X 1 + $5X 2 $280 = $7X 1 + $5X 2

13. C – 13 Graphical Solution 100 100– – 80 80 – – 60 60 – – 40 40 – – 20 20 – – ||||||||||| 020406080100 Number of Watch-TVs Number of Walkmans X1X1X1X1 X2X2X2X2 $410 = $7X 1 + $5X 2 Maximum profit line Optimal solution point (X 1 = 30, X 2 = 40)

14. C – 14 Corner-Point Method 123 100 100– – 80 80 – – 60 60 – – 40 40 – – 20 20 – – ||||||||||| 020406080100 Number of Watch-TVs Number of Walkmans X1X1X1X1 X2X2X2X2 4

15. C – 15 Solving Minimization Problems  Formulated and solved in much the same way as maximization problems  In the graphical approach an iso- cost line is used  The objective is to move the iso- cost line inwards until it reaches the lowest cost corner point

16. C – 16 Minimization Example X 1 =number of tons of black-and-white chemical produced X 2 =number of tons of color picture chemical produced Minimize total cost =2,500X 1 +3,000X 2 Subject to: X 1 ≥ 30tons of black-and-white chemical X 2 ≥ 20tons of color chemical X 1 + X 2 ≥ 60tons total X 1, X 2 ≥ $0nonnegativity requirements

17. C – 17 Minimization Example Table B.9 60 60 – 50 – 40 40 – 30 – 20 20 – 10 – – ||||||| 0102030405060 X1X1X1X1 X2X2X2X2 Feasible region X 1 = 30 X 2 = 20 X 1 + X 2 = 60 b a

18. C – 18 Minimization Example Total cost at a=2,500X 1 +3,000X 2 =2,500 (40)+3,000(20) =$160,000 Total cost at b=2,500X 1 +3,000X 2 =2,500 (30)+3,000(30) =$165,000 Lowest total cost is at point a

19. C – 19 LP Applications Production-Mix Example Department ProductWiringDrillingAssemblyInspectionUnit Profit XJ201.532.5$ 9 XM8971.5141.0$12 TR291.521.5$15 BR7881.032.5$11 CapacityMinimum Department(in hours)ProductProduction Level Wiring1,500XJ201150 Drilling2,350XM897100 Assembly2,600TR29300 Inspection1,200BR788400

20. C – 20 LP Applications X 1 = number of units of XJ201 produced X 2 = number of units of XM897 produced X 3 = number of units of TR29 produced X 4 = number of units of BR788 produced Maximize profit = 9X 1 + 12X 2 + 15X 3 + 11X 4 subject to.5X 1 +1.5X 2 +1.5X 3 +1X 4 ≤ 1,500 hours of wiring 3X 1 +1X 2 +2X 3 +3X 4 ≤ 2,350 hours of drilling 2X 1 +4X 2 +1X 3 +2X 4 ≤ 2,600 hours of assembly.5X 1 +1X 2 +.5X 3 +.5X 4 ≤ 1,200 hours of inspection X 1 ≥ 150 units of XJ201 X 1 ≥ 150 units of XJ201 X 2 ≥ 100 units of XM897 X 3 ≥ 300 units of TR29 X 4 ≥ 400 units of BR788

21. C – 21 The Simplex Method  Real world problems are too complex to be solved using the graphical method  The simplex method is an algorithm for solving more complex problems  Developed by George Dantzig in the late 1940s  Most computer-based LP packages use the simplex method

22. C – 22 NLP in Facility Location Consider an existing network with m facilitiesConsider an existing network with m facilities It is desired to add n new facilities to the networkIt is desired to add n new facilities to the network Let’sLet’s –(a i, b i ) denote the coordinates of existing facility i th –(X i, Y i ) denote the coordinates of the to-be- found new facility i th that minimize the total distribution cost

23. C – 23 NLP in Facility Location (cont.) Let’sLet’s –g ij denote the load or flow of activity from a new facility i th to an existing facility j th –f ij denote the load or flow of activity between new facilities i th and j th –c ij denote the cost per unit travel between new facilities –d ij denote the cost per unit travel between new facilities i th to an existing facility j th

24. C – 24 NLP in Facility Location (cont.) NLP ModelNLP Model

25. C – 25 Example of NLP Application บริษัทหนึ่งมีศูนย์บริการ 4 แห่ง และ Warehouse 1 แห่ง ตั้งอยู่ที่จุด Coordinate (X, Y) คือ (8,20),(8,10),(10,20),(16,30) และ (35,20) ตามลำดับ บริษัทต้องการสร้าง Warehouse อีก 2 แห่ง ซึ่งต้องตั้งห่างกันตาม แนวแกน Y และ X ไม่น้อยกว่า 5 หน่วย ปริมาณงานระหว่างนับเป็น Trip ระหว่าง Facilities มีดังตาราง และต้นทุนการขนส่ง ระหว่าง Warehouse ที่สร้างใหม่เท่ากับ 5 ต่อ ระยะทาง 1 หน่วย และ ระหว่าง Warehouse ที่ สร้างใหม่กับ Facilities เดิมเท่ากับ 10 ต่อ ระยะทาง 1 หน่วย จงกำหนดตำแหน่งที่ตั้งของ Warehouse ใหม่ทั้ง 2 แห่ง บริษัทหนึ่งมีศูนย์บริการ 4 แห่ง และ Warehouse 1 แห่ง ตั้งอยู่ที่จุด Coordinate (X, Y) คือ (8,20),(8,10),(10,20),(16,30) และ (35,20) ตามลำดับ บริษัทต้องการสร้าง Warehouse อีก 2 แห่ง ซึ่งต้องตั้งห่างกันตาม แนวแกน Y และ X ไม่น้อยกว่า 5 หน่วย ปริมาณงานระหว่างนับเป็น Trip ระหว่าง Facilities มีดังตาราง และต้นทุนการขนส่ง ระหว่าง Warehouse ที่สร้างใหม่เท่ากับ 5 ต่อ ระยะทาง 1 หน่วย และ ระหว่าง Warehouse ที่ สร้างใหม่กับ Facilities เดิมเท่ากับ 10 ต่อ ระยะทาง 1 หน่วย จงกำหนดตำแหน่งที่ตั้งของ Warehouse ใหม่ทั้ง 2 แห่ง

26. C – 26 Example of NLP Application LoadE1E2E3E4E5 WH177542 WH232452 WH1WH2 WH1 2 WH21

27. C – 27 Transportation Models

28. C – 28 Transportation Modeling  An interactive procedure that finds the least costly means of moving products from a series of sources to a series of destinations  Can be used to help resolve distribution and location decisions

29. C – 29 Transportation Modeling  A special class of linear programming  Need to know 1.The origin points and the capacity or supply per period at each 2.The destination points and the demand per period at each 3.The cost of shipping one unit from each origin to each destination

30. C – 30 Transportation Problem To FromAlbuquerqueBostonCleveland Des Moines $5$4$3 Evansville$8$4$3 Fort Lauderdale $9$7$5

31. C – 31 Transportation Problem Albuquerque (300 units required) Des Moines (100 units capacity) Evansville (300 units capacity) Fort Lauderdale (300 units capacity) Cleveland (200 units required) Boston (200 units required)

32. C – 32 Transportation Matrix From To AlbuquerqueBostonCleveland Des Moines Evansville Fort Lauderdale Factory capacity Warehouse requirement 300 200 100 700 $5 $4 $3 $9 $8 $7 Cost of shipping 1 unit from Fort Lauderdale factory to Boston warehouse Des Moines capacityconstraint Cell representing a possible source-to- destination shipping assignment (Evansville to Cleveland) Total demand and total supply Cleveland warehouse demand Figure C.2

33. C – 33 Northwest-Corner Rule  Start in the upper left-hand cell (or northwest corner) of the table and allocate units to shipping routes as follows: 1.Exhaust the supply (factory capacity) of each row before moving down to the next row 2.Exhaust the (warehouse) requirements of each column before moving to the next column 3.Check to ensure that all supplies and demands are met

34. C – 34 Northwest-Corner Rule 1.Assign 100 tubs from Des Moines to Albuquerque (exhausting Des Moines’s supply) 2.Assign 200 tubs from Evansville to Albuquerque (exhausting Albuquerque’s demand) 3.Assign 100 tubs from Evansville to Boston (exhausting Evansville’s supply) 4.Assign 100 tubs from Fort Lauderdale to Boston (exhausting Boston’s demand) 5.Assign 200 tubs from Fort Lauderdale to Cleveland (exhausting Cleveland’s demand and Fort Lauderdale’s supply)

35. C – 35 To (A) Albuquerque (B) Boston (C) Cleveland (D) Des Moines (E) Evansville (F) Fort Lauderdale Warehouse requirement 300200 Factory capacity 300 100 700 $5 $4 $3 $9 $8 $7 From Northwest-Corner Rule 100 200 Means that the firm is shipping 100 bathtubs from Fort Lauderdale to Boston

36. C – 36 Northwest-Corner Rule Computed Shipping Cost Route FromToTubs ShippedCost per UnitTotal Cost DA100$5$ 500 EA20081,600 EB1004400 FB1007700 FC2005$1,000 Total: $4,200 This is a feasible solution but not necessarily the lowest cost alternative

37. C – 37 Intuitive Lowest-Cost Method 1.Identify the cell with the lowest cost 2.Allocate as many units as possible to that cell without exceeding supply or demand; then cross out the row or column (or both) that is exhausted by this assignment 3.Find the cell with the lowest cost from the remaining cells 4.Repeat steps 2 and 3 until all units have been allocated

38. C – 38 Intuitive Lowest-Cost Method To (A) Albuquerque (B) Boston (C) Cleveland (D) Des Moines (E) Evansville (F) Fort Lauderdale Warehouse requirement 300200 Factory capacity 300 100 700 $5 $4 $3 $9 $8 $7 From 100 First, $3 is the lowest cost cell so ship 100 units from Des Moines to Cleveland and cross off the first row as Des Moines is satisfied

39. C – 39 Intuitive Lowest-Cost Method To (A) Albuquerque (B) Boston (C) Cleveland (D) Des Moines (E) Evansville (F) Fort Lauderdale Warehouse requirement 300200 Factory capacity 300 100 700 $5 $4 $3 $9 $8 $7 From 100 Second, $3 is again the lowest cost cell so ship 100 units from Evansville to Cleveland and cross off column C as Cleveland is satisfied

40. C – 40 Intuitive Lowest-Cost Method To (A) Albuquerque (B) Boston (C) Cleveland (D) Des Moines (E) Evansville (F) Fort Lauderdale Warehouse requirement 300200 Factory capacity 300 100 700 $5 $4 $3 $9 $8 $7 From 100 200 Third, $4 is the lowest cost cell so ship 200 units from Evansville to Boston and cross off column B and row E as Evansville and Boston are satisfied

41. C – 41 Intuitive Lowest-Cost Method To (A) Albuquerque (B) Boston (C) Cleveland (D) Des Moines (E) Evansville (F) Fort Lauderdale Warehouse requirement 300200 Factory capacity 300 100 700 $5 $4 $3 $9 $8 $7 From 100 200 300 Finally, ship 300 units from Albuquerque to Fort Lauderdale as this is the only remaining cell to complete the allocations

42. C – 42 Intuitive Lowest-Cost Method To (A) Albuquerque (B) Boston (C) Cleveland (D) Des Moines (E) Evansville (F) Fort Lauderdale Warehouse requirement 300200 Factory capacity 300 100 700 $5 $4 $3 $9 $8 $7 From 100 200 300 Total Cost= $3(100) + $3(100) + $4(200) + $9(300) = $4,100

43. C – 43 Intuitive Lowest-Cost Method To (A) Albuquerque (B) Boston (C) Cleveland (D) Des Moines (E) Evansville (F) Fort Lauderdale Warehouse requirement 300200 Factory capacity 300 100 700 $5 $4 $3 $9 $8 $7 From 100 200 300 Total Cost= $3(100) + $3(100) + $4(200) + $9(300) = $4,100 This is a feasible solution, and an improvement over the previous solution, but not necessarily the lowest cost alternative